Recursive sequence depending on the parameter

Question

For the given parameter $mathbb Rni tgeq 1$, the sequence is
  defined recursively: $$a_1=t,;;a_{n+1}a_n=3a_n-2$$ $(a)$ Let $t=4$.
  Prove the sequence $(a_n)$ converges and find its limit.
  
  $(b)$ Which parameters $tgeq 1$ is the sequence $(a_n)$ increasing
  for?

My attempt:

Bolzano-Weierstrass:A sequence converges if it is monotonous and
   bounded

$$a_{n+1}a_n=3a_n-2implies a_{n+1}=3-frac{2}{a_n}$$
$(a)$

First few terms: $a_1=4,a_2=frac{5}{2},a_3=frac{11}{5}$

Assumption: the sequence is decreasing

Proof by induction:
the basis (n=1) is trivial: $frac{5}{2}<4$

Assumption: $a_n<a_{n-1},;forall ninmathbb N$

Step: $$a_n<a_{n-1}impliesfrac{1}{a_n}geqfrac{1}{a_{n-1}}Bigg/cdot(-2)$$
$$iff-frac{2}{a_n}leq-frac{2}{a_{n-1}}iff underbrace{3-frac{2}{a_n}}_{a_{n+1}}lequnderbrace{3-frac{2}{a_{n-1}}}_{a_n}$$
The limit: $$L=3-frac{2}{L}implies L^2-3L+2=0$$
I take into account only $2$ because the parabola is convex and $$a_nto L^-.$$
Then I have to prove: $a_ngeq 2;forall ninmathbb N$ after the formal computing: $a_{n+1}geq 3-frac{2}{2}=2$
$underset{implies}{text{Bolzano-Weierstrass theorem}}(a_n)to 2$

$(b)$ Since the sequence doesn't have to be convergent, only increasing:
$$a_2=3-frac{2}{t}geq timplies tin[1,2]$$
Then, it should follow inductively,analogously to $(a)$, this time it is increasing.
Is this correct?

dan_fulea · Accepted Answer

Let us find the general form for the sequence given by the recursion
$$
a_{n+1}=
underbrace{
begin{bmatrix}3&-2\ 1&0
end{bmatrix}
}_{}cdot a_n ,
$$
where we use the Möbius action of matrices $2times 2$ on scalars, given in general by 
$$
begin{bmatrix}a&b\ c&d
end{bmatrix}cdot x
:=
frac{ax+b}{cx+d} ,
$$
see also Möbius transformation, wiki page.

The special matrix $A$ used in the problem can be diagonalized,
$$
A=
underbrace{
begin{bmatrix}1&1\ 1/2&1
end{bmatrix}}_{T}
underbrace{
begin{bmatrix}2&\ &1
end{bmatrix}}_{D}
underbrace{
begin{bmatrix}2&-2\-1&2
end{bmatrix}}_{T^{-1}}
$$ 
and because $A^n=TD^nT^{-1}$ we get the general form for $a_n=A^ncdotbegin{bmatrix}4\1end{bmatrix}$, and then passing to the element in the projective space, by taking the quotient, it is:
$$
a_n=frac{6cdot 2^n-2}{3cdot 2^n-2} .
$$
It converges to $6/3=2$.

For the part (b) a similar study can be started, the general term being
$$
a_n(t)=
TD^nT^{-1}
begin{bmatrix}t\ 1
end{bmatrix}_{Bbb P^1}
=
begin{bmatrix}
2cdot 2^n-1 & -2cdot 2^n+2\
2^n-1 & -2cdot 2^n+2
end{bmatrix}
begin{bmatrix}
t\ 1
end{bmatrix}
text{ considered in }
{Bbb P^1} 
 .
$$
I am stopping here...

Neat Math · Answer

If you have already proven $a_n > 2, forall n$ , then
Let $b_n=a_n-2$, $b_n>0$. $a_{n+1} = 3 - frac{2}{a_n} Rightarrow b_{n+1} + 2=3-frac{2}{b_n+2} Rightarrow b_{n+1} = frac{b_n}{2+b_n} < frac{b_n}{2}$.
Therefore as $nto infty, b_n to 0, a_n to 2.blacksquare$
The next is pure hindsight based on dan_fulea's solution but I believe it can be useful when the two fixed points are distinct.
$a_{n+1} - 1 = 2-frac{2}{a_n} = frac{2(a_n - 1)}{a_n}$
$a_{n+1} - 2 = 1-frac{2}{a_n} = frac{a_n - 2}{a_n}$
Therefore $frac{a_{n+1}-1}{a_{n+1}-2} = 2cdot frac{a_n-1}{a_n-2}$
$frac{a_n-1}{a_n-2}$ is a geometric sequence with initial value $frac{3}{2}$,
so $frac{a_n-1}{a_n-2} = 2^{n-1} frac{3}{2} = 1+frac{1}{a_n-2} Rightarrow a_n = 2+ frac{1}{2^{n-1}frac{3}{2}-1} = frac{6cdot 2^{n-1}-2}{3cdot 2^{n-1}-2}.blacksquare$
In general if there are two distinct fixed points $r$ and $s$ then the ratio $frac{a_n-r}{a_n-s}$ is a geometric sequence.

Recursive sequence depending on the parameter

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